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aliina [53]
2 years ago
5

A distant astronomical object (a quasar) is moving away from us at half the speed of light. what is the speed of the light we re

ceive from this quasar?
Physics
1 answer:
Aloiza [94]2 years ago
7 0

For a distant astronomical object (a quasar) is moving away from us at half the speed of light, the speed of the light we receive from this quasar  is mathematically given as c = 3x108 m/s

<h3>What is the speed of light?</h3>

Generally, the equation for the   is mathematically given as

The speed of light can be said to be measured to be approximately the value c = 3x108 m/s

.

In conclusion, An will not influence  3x108 m/s unless both are inertia and frames refrences where newtons laws are valid, so the speed of light measured from the earth frame is equal to  3x108 m/s

Read more about Speed

brainly.com/question/4931057

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The driver of a car travels at 90 km / h, observes some children playing on the road 50 m away, and applies the brakes, managing
Orlov [11]

Answer:

13,750 N

Yes

Explanation:

Given:

v₀ = 90 km/h = 25 m/s

v = 0 m/s

t = 4 s

Find: a and Δx

a = Δv / Δt

a = (0 m/s − 25 m/s) / (4 s)

a = -6.25 m/s²

F = ma

F = (2200 kg) (-6.25 m/s²)

F = -13,750 N

Δx = ½ (v + v₀) t

Δx = ½ (0 m/s + 25 m/s) (4 s)

Δx = 50 m

6 0
3 years ago
A boxer hits punching bag and gives it a change in momentum of 12 kg multiplied by m divided by s over 7.0ms what is the magnitu
mr_godi [17]

Answer: 1700

Explanation:

8 0
2 years ago
A 1kg mass is thrown to a height of 2cm. what is the potential energy​
prohojiy [21]
  • Mass=m=1kg
  • Height=h=2cm=0.02m
  • Acceleration due to gravity=g=10m/s^2

\\ \tt\hookrightarrow P.E=mgh

\\ \tt\hookrightarrow PE=1(10)(0.02)

\\ \tt\hookrightarrow PE=0.2J

4 0
2 years ago
Read 2 more answers
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0
3 years ago
If two vectors are perpendicular to each other, how should you add them?
Aleksandr [31]

Let us consider two vectors A and B.

As per the question, the two vectors are perpendicular to each other.

Hence the angle between them  \theta =90 degree

We are asked to calculate the resultant of these two vectors.

As per parallelogram law of vector addition, the resultant of two vectors are-

                      R=\sqrt{A^{2}+ B^{2}+2ABcos\theta

                                =\sqrt{ A^{2}+ B^{2}+2AB*cos90}    [cos90=0]

                                =\sqrt{ A^{2}+ B^{2}

This is the way by which we can add two perpendicular vectors.


8 0
3 years ago
Read 2 more answers
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